One of the most powerful ideas in physical chemistry is that the same thermodynamic quantity, the standard Gibbs energy change, connects chemical equilibrium and electrochemistry.
From chemical equilibrium, we learned that
\[ \Delta G^\circ = -RT\ln K \]where \(K\) is the thermodynamic equilibrium constant.
In electrochemistry, we found that the maximum electrical work produced by an electrochemical cell is related to the Gibbs energy change through
\[ \Delta G^\circ = -nFE^\circ \]where
Since both expressions describe the same thermodynamic quantity, they can be equated.
Setting the two expressions for the standard Gibbs energy equal gives
\[ -nFE^\circ = -RT\ln K \]Rearranging,
\[ E^\circ = \frac{RT}{nF} \ln K \]At
\[ 298\ \text{K} \]this becomes
\[ E^\circ = \frac{0.02569}{n} \ln K \]or, using common logarithms,
\[ E^\circ = \frac{0.05916}{n} \log K \]This equation allows equilibrium constants to be determined directly from standard cell potentials.
| \(E^\circ\) | \(\Delta G^\circ\) | \(K\) | Reaction |
|---|---|---|---|
| Positive | Negative | \(K > 1\) | Products favored |
| Zero | Zero | \(K = 1\) | Neither direction favored |
| Negative | Positive | \(K < 1\) | Reactants favored |
Thus, a large positive standard cell potential corresponds to a spontaneous reaction with a large equilibrium constant, while a negative standard cell potential corresponds to a non-spontaneous reaction and a small equilibrium constant.
Electrochemical measurements provide a convenient experimental method for determining equilibrium constants. Measuring a single quantity, the standard cell potential, allows us to determine both the standard Gibbs energy change and the equilibrium constant for a reaction.
Conversely, if the equilibrium constant is known from equilibrium measurements, the standard cell potential can be predicted without constructing an electrochemical cell.
Big picture: The relationships \[ \Delta G^\circ=-nFE^\circ \] and \[ \Delta G^\circ=-RT\ln K \] show that electrochemistry and chemical equilibrium are simply two different ways of describing the same thermodynamic driving force. Cell potentials, equilibrium constants, and Gibbs energies are all different measures of the tendency of a chemical reaction to proceed.
Calculate \(E^\circ\), \(\Delta G^\circ\), and \(K\) at 298 K for the reaction
\[ \mathrm{Cu(s)+2Ag^+(aq)\rightleftharpoons Cu^{2+}(aq)+2Ag(s)} \]Use the following standard reduction potentials:
| Reduction half-reaction | \(E^\circ_{\text{red}}\) |
|---|---|
| \(\mathrm{Ag^+(aq)+e^- \rightarrow Ag(s)}\) | \(+0.799\ \text{V}\) |
| \(\mathrm{Cu^{2+}(aq)+2e^- \rightarrow Cu(s)}\) | \(+0.337\ \text{V}\) |
Silver ion has the larger standard reduction potential, so silver is reduced at the cathode:
\[ \mathrm{Ag^+(aq)+e^- \rightarrow Ag(s)} \]Copper is oxidized at the anode:
\[ \mathrm{Cu(s)\rightarrow Cu^{2+}(aq)+2e^-} \]The standard cell potential is
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] \[ E^\circ_{\text{cell}} = 0.799-0.337 = 0.462\ \text{V} \]Therefore,
\[ \boxed{E^\circ=0.462\ \text{V}} \]For the balanced reaction, two moles of electrons are transferred:
\[ n=2 \]The standard Gibbs energy change is related to the standard cell potential by
\[ \Delta G^\circ=-nFE^\circ \] \[ \Delta G^\circ = -(2)(96485\ \text{C mol}^{-1})(0.462\ \text{J C}^{-1}) \] \[ \Delta G^\circ = -8.92\times10^4\ \text{J mol}^{-1} \] \[ \boxed{\Delta G^\circ=-89.2\ \text{kJ mol}^{-1}} \]Finally, use
\[ \Delta G^\circ=-RT\ln K \]so
\[ \ln K = -\frac{\Delta G^\circ}{RT} \] \[ \ln K = \frac{8.92\times10^4} {(8.314)(298)} = 36.0 \] \[ K=e^{36.0} = 4.3\times10^{15} \] \[ \boxed{K=4.3\times10^{15}} \]Physical interpretation: The positive value of \(E^\circ\) means the reaction is spontaneous under standard conditions. This corresponds to a negative \(\Delta G^\circ\) and a very large equilibrium constant, meaning that the reaction strongly favors formation of \(\mathrm{Cu^{2+}}\) and \(\mathrm{Ag(s)}\).
Two standard reduction half-reactions are shown below. Use them to determine the spontaneous redox reaction, then calculate \(E^\circ\), \(\Delta G^\circ\), and \(K\) at 298 K.
| Half-reaction | \(E^\circ_{\text{red}}\) |
|---|---|
The half-reaction with the larger reduction potential will occur as the reduction. The other half-reaction will be reversed to give the oxidation.
| Quantity | Your answer |
|---|---|
| \(E^\circ_{\text{cell}}\) | V |
| \(\Delta G^\circ\) | kJ mol-1 |
| \(K\) | |
Big picture: The relationships \[ \Delta G^\circ=-nFE^\circ \] and \[ \Delta G^\circ=-RT\ln K \] unify electrochemistry, thermodynamics, and chemical equilibrium. Standard cell potentials, Gibbs energies, and equilibrium constants are simply different ways of describing the same thermodynamic driving force behind a chemical reaction.